Recall that in solving simple first order "ordinary differential equation" (ODE), we may apply variable separable differential equation wherein:
`N(y)y'=M(x)`
`N(y)(dy)/(dx)=M(x)`
`N(y) dy=M(x) dx`
Before we can work on the direct integration:` ` int N(y) dy= int M(x) dx to solve for the general solution of a differential equation.
For the given first order ODE: `(dy)/(dx)=y+3 ` can be rearrange by cross-multiplication into:
`(dy)/(y+3)=dx`
Apply direct integration on both sides:` int(dy)/(y+3)=int ` dx
For the left side, we consider u-substitution by letting:
`u= y+3` then` du = dy`
The integral becomes:
`int(dy)/(y+3)=int(du)/(u)`
Applying basic integration formula for logarithm:
`int(du)/(u)= ln|u|`
Plug-in `u = y+3` on `ln|u`` |` , we get:
`int(dy)/(y+3)=ln|y+3|`
For the right side, we apply the basic integration: `int dx= x+C`
Combing the results from both sides, we get the general solution of the differential equation as:
`ln|y+3|= x+C`
or
`y =e^(x+C)-3`
`y =Ce^x-3`
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